3.504 \(\int (d+c d x)^{5/2} \sqrt{f-c f x} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=376 \[ \frac{1}{4} c^2 d^2 x^3 \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 d^2 \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}-\frac{2 d^2 \left (1-c^2 x^2\right ) \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{3}{8} d^2 x \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac{b c^3 d^2 x^4 \sqrt{c d x+d} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}-\frac{2 b c^2 d^2 x^3 \sqrt{c d x+d} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{3 b c d^2 x^2 \sqrt{c d x+d} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}+\frac{2 b d^2 x \sqrt{c d x+d} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}} \]

[Out]

(2*b*d^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (3*b*c*d^2*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*
f*x])/(16*Sqrt[1 - c^2*x^2]) - (2*b*c^2*d^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) - (b*c^
3*d^2*x^4*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(16*Sqrt[1 - c^2*x^2]) + (3*d^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(
a + b*ArcSin[c*x]))/8 + (c^2*d^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/4 - (2*d^2*Sqrt[d +
c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c) + (5*d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a +
b*ArcSin[c*x])^2)/(16*b*c*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.536966, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4673, 4763, 4647, 4641, 30, 4677, 4697, 4707} \[ \frac{1}{4} c^2 d^2 x^3 \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 d^2 \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}-\frac{2 d^2 \left (1-c^2 x^2\right ) \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{3}{8} d^2 x \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac{b c^3 d^2 x^4 \sqrt{c d x+d} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}-\frac{2 b c^2 d^2 x^3 \sqrt{c d x+d} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{3 b c d^2 x^2 \sqrt{c d x+d} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}+\frac{2 b d^2 x \sqrt{c d x+d} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + c*d*x)^(5/2)*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]),x]

[Out]

(2*b*d^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (3*b*c*d^2*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*
f*x])/(16*Sqrt[1 - c^2*x^2]) - (2*b*c^2*d^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) - (b*c^
3*d^2*x^4*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(16*Sqrt[1 - c^2*x^2]) + (3*d^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(
a + b*ArcSin[c*x]))/8 + (c^2*d^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/4 - (2*d^2*Sqrt[d +
c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c) + (5*d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a +
b*ArcSin[c*x])^2)/(16*b*c*Sqrt[1 - c^2*x^2])

Rule 4673

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[((d + e*x)^q*(f + g*x)^q)/(1 - c^2*x^2)^q, Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4697

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((
f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1 -
c^2*x^2]), Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*(m
+ 2)*Sqrt[1 - c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}
, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

\begin{align*} \int (d+c d x)^{5/2} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{\left (\sqrt{d+c d x} \sqrt{f-c f x}\right ) \int (d+c d x)^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (\sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \left (d^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+2 c d^2 x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+c^2 d^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}+\frac{\left (2 c d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}+\frac{\left (c^2 d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{1}{2} d^2 x \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} c^2 d^2 x^3 \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c d x} \sqrt{f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{\left (d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{2 \sqrt{1-c^2 x^2}}+\frac{\left (2 b d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \left (1-c^2 x^2\right ) \, dx}{3 \sqrt{1-c^2 x^2}}-\frac{\left (b c d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x \, dx}{2 \sqrt{1-c^2 x^2}}+\frac{\left (c^2 d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{4 \sqrt{1-c^2 x^2}}-\frac{\left (b c^3 d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x^3 \, dx}{4 \sqrt{1-c^2 x^2}}\\ &=\frac{2 b d^2 x \sqrt{d+c d x} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}}-\frac{b c d^2 x^2 \sqrt{d+c d x} \sqrt{f-c f x}}{4 \sqrt{1-c^2 x^2}}-\frac{2 b c^2 d^2 x^3 \sqrt{d+c d x} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{b c^3 d^2 x^4 \sqrt{d+c d x} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}+\frac{3}{8} d^2 x \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} c^2 d^2 x^3 \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c d x} \sqrt{f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{d^2 \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}+\frac{\left (d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{8 \sqrt{1-c^2 x^2}}+\frac{\left (b c d^2 \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x \, dx}{8 \sqrt{1-c^2 x^2}}\\ &=\frac{2 b d^2 x \sqrt{d+c d x} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}}-\frac{3 b c d^2 x^2 \sqrt{d+c d x} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}-\frac{2 b c^2 d^2 x^3 \sqrt{d+c d x} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{b c^3 d^2 x^4 \sqrt{d+c d x} \sqrt{f-c f x}}{16 \sqrt{1-c^2 x^2}}+\frac{3}{8} d^2 x \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} c^2 d^2 x^3 \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c d x} \sqrt{f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{5 d^2 \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.23363, size = 293, normalized size = 0.78 \[ \frac{d^2 \sqrt{c d x+d} \sqrt{f-c f x} \left (48 a \sqrt{1-c^2 x^2} \left (6 c^3 x^3+16 c^2 x^2+9 c x-16\right )-256 b c x \left (c^2 x^2-3\right )+144 b \cos \left (2 \sin ^{-1}(c x)\right )-9 b \cos \left (4 \sin ^{-1}(c x)\right )\right )-720 a d^{5/2} \sqrt{f} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )+12 b d^2 \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x) \left (-64 \left (1-c^2 x^2\right )^{3/2}+24 \sin \left (2 \sin ^{-1}(c x)\right )-3 \sin \left (4 \sin ^{-1}(c x)\right )\right )+360 b d^2 \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)^2}{1152 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c*d*x)^(5/2)*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]),x]

[Out]

(360*b*d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*a*d^(5/2)*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x
*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-25
6*b*c*x*(-3 + c^2*x^2) + 48*a*Sqrt[1 - c^2*x^2]*(-16 + 9*c*x + 16*c^2*x^2 + 6*c^3*x^3) + 144*b*Cos[2*ArcSin[c*
x]] - 9*b*Cos[4*ArcSin[c*x]]) + 12*b*d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-64*(1 - c^2*x^2)^(3/2)
+ 24*Sin[2*ArcSin[c*x]] - 3*Sin[4*ArcSin[c*x]]))/(1152*c*Sqrt[1 - c^2*x^2])

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Maple [F]  time = 0.338, size = 0, normalized size = 0. \begin{align*} \int \left ( cdx+d \right ) ^{{\frac{5}{2}}} \left ( a+b\arcsin \left ( cx \right ) \right ) \sqrt{-cfx+f}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^(5/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x)

[Out]

int((c*d*x+d)^(5/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(5/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a c^{2} d^{2} x^{2} + 2 \, a c d^{2} x + a d^{2} +{\left (b c^{2} d^{2} x^{2} + 2 \, b c d^{2} x + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{c d x + d} \sqrt{-c f x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(5/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="fricas")

[Out]

integral((a*c^2*d^2*x^2 + 2*a*c*d^2*x + a*d^2 + (b*c^2*d^2*x^2 + 2*b*c*d^2*x + b*d^2)*arcsin(c*x))*sqrt(c*d*x
+ d)*sqrt(-c*f*x + f), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**(5/2)*(a+b*asin(c*x))*(-c*f*x+f)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c d x + d\right )}^{\frac{5}{2}} \sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(5/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="giac")

[Out]

integrate((c*d*x + d)^(5/2)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a), x)